Let
- $(\Omega,\mathcal A)$ be a measurable space
- $\mu$ be a finite measure on $(\Omega,\mathcal A)$
- $\kappa$ be a sub-Markov kernel on $(\Omega,\mathcal A)$
- $p\ge1$
I'll denote the composition of $\mu$ and $\kappa$ by $\mu\kappa$. Moreover, $$\kappa f:=\int\kappa(x,{\rm d}y)f(y)$$ and $$\mu f:=\int f\:{\rm d}\mu$$ whenever the integrals are well-defined. It's easy to see that if $f:\Omega\to\overline{\mathbb R}$ is $\mathcal A$-measurable and either nonnegative or bounded, then $\kappa f$ is $\mathcal A$-measurable.
Now, assume $$\mu\kappa\le\mu\tag1$$ and let $g\in\mathcal L^p(\mu)$. I want to show that $\kappa g\in L^p(\mu)$ and $$\left\|\kappa g\right\|_{L^p(\mu)}\le\left\|g\right\|_{L^p(\mu)}\tag2.$$
By the aforementioned property, $\kappa|g|$ is well-defined and $\mathcal A$-measurable. Moreover, by a Fubini like result $$\mu(\kappa f)=(\mu\kappa)f\tag3$$ for all $\mathcal A$-measurable $f:\Omega\to[0,\infty]$ and hence $$\mu(\kappa|g|)=(\mu\kappa)|g|<\infty.$$ Thus, $$\kappa|g|<\infty\;\;\;\mu\text{-almost everywhere}\tag4$$ and hence $\kappa g$ is $\mu$-almost everywhere well-defined. We can find a sequence $(g_n)_{n\in\mathbb N}$ of (elementary, simple, step - call it however you like) $\mathcal A$-measurable functions with $$g_n\xrightarrow{n\to\infty}g\tag5$$ and $$\left|g_n\right|\le\left|g\right|\;\;\;\text{for all }n\in\mathbb N\tag6.$$ Thus, by dominated convergence, $$\kappa g_n\xrightarrow{n\to\infty}\kappa g\;\;\;\mu\text{-almost everywhere}\tag7.$$ So, there is a $\mathcal A$-measurable $f:\Omega\to\mathbb R$ (e.g. $f:=\limsup_{n\to\infty}\kappa g_n$) with $$f=\kappa g\;\;\;\mu\text{-almost everywhere}\tag8.$$ By Jensen's inequality $$\left|\kappa g\right|^p\le\kappa|g|^p$$ and hence $$\left\|f\right\|_{L^p(\mu)}^p\le\mu(\kappa|g|^p)=(\mu\kappa)|g|^p=\left\|g\right\|_{L^p(\mu)}^p\tag9.$$
Did I overcomplicate anything? Please let me know. In any case, my actual question is: Is it correct that we're not able to show that $\kappa g$ is well-defined and $\mathcal A$-measurable on all of $\Omega$? As shown in my proof, I guess that it's only possible to show that there is a $\mathcal A$-measurable $\mu$-version of $\kappa g$. Or am I missing something?